
We can show that the edges $AD$ and $BC$ of
a tetrahedron $ABCD$ are mutually perpendicular if and only if
$AB^2 + CD^2 = AC^2 + BD^2$. A tetrahedron has three pairs of
opposite edges. We have to prove that one pair are mutually
perpendicular if and only if the sums of the squares of the lengths
of the other two pairs are equal.
Edward of Graveney School, Tooting, London
sent in an excellent solution to this question using vectors.

It is easiest to follow if the essential symmetry is made
obvious by the notation. In order to do this we use the position
vectors ${\bf a, b, c}$ and ${\bf d}$ for the vertices of the
tetrahedron. The scalar product ${\bf a.a} = a^2$ gives the square
of the length of the vector ${\bf a}$.
We know that
$$\eqalign { AB^2 + CD^2 &= ({\bf b  a})^2 + ({\bf d 
c})^2 \cr &= a^2 + b^2 + c^2 + d^2  2 {\bf b.a}  2{\bf
d.c}}$$
and
$$\eqalign { AC^2 + BD^2 &= ({\bf c  a})^2 + ({\bf d 
b})^2 \cr &= a^2 + b^2 + c^2 + d^2  2 {\bf c.a}  2{\bf
d.b}.}$$
Therefore
$$\eqalign { (AB^2 + CD^2) (AC^2 + BD^2) &= 2({\bf c.a +
d.b  b.a  d.c}) \cr &= 2({\bf c  b}).({\bf a  d}) }$$
The right hand side of this expression is just twice the
scalar product of the vectors $\mathbf{BC}$ and $\mathbf{DA}$ and
we know that the edges $BC$ and $DA$ are mutually perpendicular if
and only if this scalar product is zero. The result follows.